Description#
You are installing a billboard and want it to have the largest height. The billboard will have two steel supports, one on each side. Each steel support must be an equal height.
You are given a collection of rods that can be welded together. For example, if you have rods of lengths 1, 2, and 3, you can weld them together to make a support of length 6.
Return the largest possible height of your billboard installation. If you cannot support the billboard, return 0.
Example 1:
Input: rods = [1,2,3,6]
Output: 6
Explanation: We have two disjoint subsets {1,2,3} and {6}, which have the same sum = 6.
Example 2:
Input: rods = [1,2,3,4,5,6]
Output: 10
Explanation: We have two disjoint subsets {2,3,5} and {4,6}, which have the same sum = 10.
Example 3:
Input: rods = [1,2]
Output: 0
Explanation: The billboard cannot be supported, so we return 0.
Constraints:
1 <= rods.length <= 201 <= rods[i] <= 1000sum(rods[i]) <= 5000
Solutions#
Solution 1#
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| class Solution:
def tallestBillboard(self, rods: List[int]) -> int:
@cache
def dfs(i: int, j: int) -> int:
if i >= len(rods):
return 0 if j == 0 else -inf
ans = max(dfs(i + 1, j), dfs(i + 1, j + rods[i]))
ans = max(ans, dfs(i + 1, abs(rods[i] - j)) + min(j, rods[i]))
return ans
return dfs(0, 0)
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| class Solution {
private Integer[][] f;
private int[] rods;
private int n;
public int tallestBillboard(int[] rods) {
int s = 0;
for (int x : rods) {
s += x;
}
n = rods.length;
this.rods = rods;
f = new Integer[n][s + 1];
return dfs(0, 0);
}
private int dfs(int i, int j) {
if (i >= n) {
return j == 0 ? 0 : -(1 << 30);
}
if (f[i][j] != null) {
return f[i][j];
}
int ans = Math.max(dfs(i + 1, j), dfs(i + 1, j + rods[i]));
ans = Math.max(ans, dfs(i + 1, Math.abs(rods[i] - j)) + Math.min(j, rods[i]));
return f[i][j] = ans;
}
}
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| class Solution {
public:
int tallestBillboard(vector<int>& rods) {
int s = accumulate(rods.begin(), rods.end(), 0);
int n = rods.size();
int f[n][s + 1];
memset(f, -1, sizeof(f));
function<int(int, int)> dfs = [&](int i, int j) -> int {
if (i >= n) {
return j == 0 ? 0 : -(1 << 30);
}
if (f[i][j] != -1) {
return f[i][j];
}
int ans = max(dfs(i + 1, j), dfs(i + 1, j + rods[i]));
ans = max(ans, dfs(i + 1, abs(j - rods[i])) + min(j, rods[i]));
return f[i][j] = ans;
};
return dfs(0, 0);
}
};
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| func tallestBillboard(rods []int) int {
s := 0
for _, x := range rods {
s += x
}
n := len(rods)
f := make([][]int, n)
for i := range f {
f[i] = make([]int, s+1)
for j := range f[i] {
f[i][j] = -1
}
}
var dfs func(i, j int) int
dfs = func(i, j int) int {
if i >= n {
if j == 0 {
return 0
}
return -(1 << 30)
}
if f[i][j] != -1 {
return f[i][j]
}
ans := max(dfs(i+1, j), dfs(i+1, j+rods[i]))
ans = max(ans, dfs(i+1, abs(j-rods[i]))+min(j, rods[i]))
f[i][j] = ans
return ans
}
return dfs(0, 0)
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
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| function tallestBillboard(rods: number[]): number {
const s = rods.reduce((a, b) => a + b, 0);
const n = rods.length;
const f = new Array(n).fill(0).map(() => new Array(s + 1).fill(-1));
const dfs = (i: number, j: number): number => {
if (i >= n) {
return j === 0 ? 0 : -(1 << 30);
}
if (f[i][j] !== -1) {
return f[i][j];
}
let ans = Math.max(dfs(i + 1, j), dfs(i + 1, j + rods[i]));
ans = Math.max(ans, dfs(i + 1, Math.abs(j - rods[i])) + Math.min(j, rods[i]));
return (f[i][j] = ans);
};
return dfs(0, 0);
}
|
Solution 2#
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| class Solution:
def tallestBillboard(self, rods: List[int]) -> int:
n = len(rods)
s = sum(rods)
f = [[-inf] * (s + 1) for _ in range(n + 1)]
f[0][0] = 0
t = 0
for i, x in enumerate(rods, 1):
t += x
for j in range(t + 1):
f[i][j] = f[i - 1][j]
if j >= x:
f[i][j] = max(f[i][j], f[i - 1][j - x])
if j + x <= t:
f[i][j] = max(f[i][j], f[i - 1][j + x] + x)
if j < x:
f[i][j] = max(f[i][j], f[i - 1][x - j] + x - j)
return f[n][0]
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| class Solution {
public int tallestBillboard(int[] rods) {
int n = rods.length;
int s = 0;
for (int x : rods) {
s += x;
}
int[][] f = new int[n + 1][s + 1];
for (var e : f) {
Arrays.fill(e, -(1 << 30));
}
f[0][0] = 0;
for (int i = 1, t = 0; i <= n; ++i) {
int x = rods[i - 1];
t += x;
for (int j = 0; j <= t; ++j) {
f[i][j] = f[i - 1][j];
if (j >= x) {
f[i][j] = Math.max(f[i][j], f[i - 1][j - x]);
}
if (j + x <= t) {
f[i][j] = Math.max(f[i][j], f[i - 1][j + x] + x);
}
if (j < x) {
f[i][j] = Math.max(f[i][j], f[i - 1][x - j] + x - j);
}
}
}
return f[n][0];
}
}
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| class Solution {
public:
int tallestBillboard(vector<int>& rods) {
int n = rods.size();
int s = accumulate(rods.begin(), rods.end(), 0);
int f[n + 1][s + 1];
memset(f, -0x3f, sizeof(f));
f[0][0] = 0;
for (int i = 1, t = 0; i <= n; ++i) {
int x = rods[i - 1];
t += x;
for (int j = 0; j <= t; ++j) {
f[i][j] = f[i - 1][j];
if (j >= x) {
f[i][j] = max(f[i][j], f[i - 1][j - x]);
}
if (j + x <= t) {
f[i][j] = max(f[i][j], f[i - 1][j + x] + x);
}
if (j < x) {
f[i][j] = max(f[i][j], f[i - 1][x - j] + x - j);
}
}
}
return f[n][0];
}
};
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| func tallestBillboard(rods []int) int {
n := len(rods)
s := 0
for _, x := range rods {
s += x
}
f := make([][]int, n+1)
for i := range f {
f[i] = make([]int, s+1)
for j := range f[i] {
f[i][j] = -(1 << 30)
}
}
f[0][0] = 0
for i, t := 1, 0; i <= n; i++ {
x := rods[i-1]
t += x
for j := 0; j <= t; j++ {
f[i][j] = f[i-1][j]
if j >= x {
f[i][j] = max(f[i][j], f[i-1][j-x])
}
if j+x <= t {
f[i][j] = max(f[i][j], f[i-1][j+x]+x)
}
if j < x {
f[i][j] = max(f[i][j], f[i-1][x-j]+x-j)
}
}
}
return f[n][0]
}
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